If two operators are related to each other by a unitary operator, are their eigenvalues the same?

Question

If we consider 2 linearly independent basis as follow: $$\{ |\psi_1 \rangle , |\psi_2 \rangle ... |\psi_n \rangle\}$$ $$\{ |\phi_1 \rangle , |\phi_2 \rangle ... |\phi_n \rangle\}$$

And they are related by a unitary tranform such that: $$U|\psi_i\rangle = |\phi_n\rangle$$

If O is an oprator in basis $\{ |\psi_1 \rangle , |\psi_2 \rangle ... |\psi_n \rangle\}$ and O' is operator in basis $\{ |\phi_1 \rangle , |\phi_2 \rangle ... |\phi_n \rangle\}$, and we could write the spectral decomposition as follows:

$$O = \sum_n \lambda_n |\psi_n \rangle \langle\psi_n|$$ $$O' = \sum_n \mu_n |\phi_n \rangle \langle\phi_n|$$

Would it be true to say that the eigenvalues $\mu_n$ and $\lambda_n$ are equal to one another?

Answer 1

You probably mean that, since $\langle \psi_n|U^\dagger =\langle \phi_n|$, $$ O'=UOU^\dagger = \sum_n \lambda_n U|\psi_n \rangle \langle\psi_n| U^\dagger= \sum_n \lambda_n |\phi_n \rangle \langle\phi_n|= O' ,$$ unitarily equivalent.

So, if $O$ and $O'$ are unitarily equivalent, the corresponding transformation of their eigenstate decompositions forces $\lambda_n=\mu_n$.

If they are not unitarily equivalent, the eigenvalues won't coincide: any operator can be written in any unitarily equivalent basis, and the basis by itself cannot tell it what it is or isn't!

Try a 2x2 matrix example. Note the trace and determinant of two unitarily equivalent matrices coincide.

Answer 2

It is given that $$U|\psi_n\rangle = |\phi_n\rangle$$ where $U$ is a unitary matrix and $|\psi_n\rangle$ are eigenkets of $\hat{O}$ and $|\phi_n\rangle$ are eigenkets of $\hat{O'}$ with eigenvalues $\lambda_n$ and $\mu_n$, respectively.

Equivalently, $$|\psi_n\rangle = U^\dagger|\phi_n\rangle$$

Since $\hat{O'}|\phi_n\rangle = \mu_n|\phi_n\rangle$ and $\hat{O}|\psi_n\rangle = \lambda_n |\psi_n\rangle$, and since $\hat{O'} = (U\hat{O}U^\dagger)$, then

$$ \mu_n|\phi_n\rangle = \hat{O'}|\phi_n\rangle = \hat{O'} U|\psi_n\rangle = (U\hat{O}U^\dagger)U|\psi_n\rangle = U\hat{O}|\psi_n\rangle = U\lambda_n|\psi_n\rangle = \lambda_n U|\psi_n\rangle = \lambda_n |\phi_n\rangle$$

Thus, $\lambda_n = \mu_n$ and they share the same eigenvalues.